p-is-apolynomial-having-n-roots-x-i-with-x-i-xj-for-i-j-calculate-k-1-n-1-1-x-k-

Question Number 32334 by abdo imad last updated on 23/Mar/18

p i s a p o l y n o m i a l h a v i n g n r o o t s (x_{i}) w i t h x_{i} \neq x j f o r

i \neq j c a l c u l a t e \sum_{k = 1}^{n} \frac{1}{1 - x_{k}} .

Commented by abdo imad last updated on 01/Apr/18

p (x) = λ \prod_{k = 1}^{n} (x - x_{i}) \Rightarrow p^{^{'}} (x) = λ \sum_{k = 1}^{n} \prod_{p = 1_{p \neq k}}^{n} (x - x_{p})

\Rightarrow \frac{p^{^{'}} (x)}{p (x)} = \sum_{k = 1}^{n} \frac{1}{x - x_{k}} \Rightarrow \sum_{k = 1}^{n} \frac{1}{1 - x_{k}} = \frac{p^{^{'}} (1)}{p (1)} w i t h p (1) \neq 0